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Gromov-Hausdorff distances from simply connected geodesic spaces to the circle.

Saúl Rodríguez Martín The Ohio State University, [email protected]
Abstract

We prove that the Gromov-Hausdorff distance from the circle with its geodesic metric to any simply connected geodesic space is never smaller than π4\frac{\pi}{4}. We also prove that this bound is tight through the construction of a simply connected geodesic space E\mathrm{E} which attains the lower bound π4\frac{\pi}{4}. We deduce the first statement from a general result that we also establish which gives conditions on how small the Gromov-Hausdorff distance between two geodesic metric spaces (X,dX)(X,d_{X}) and (Y,dY)(Y,d_{Y}) has to be in order for π1(X)\pi_{1}(X) and π1(Y)\pi_{1}(Y) to be isomorphic.

1 Introduction

In the following, for n1n\geq 1 we give the unit sphere 𝕊n:={zn+1;z=1}\mathbb{S}^{n}:=\{z\in\mathbb{R}^{n+1};\|z\|=1\} its intrinsic metric d𝕊nd_{\mathbb{S}^{n}}. That is, for any two points p,q𝕊np,q\in\mathbb{S}^{n} seen as vectors in n+1\mathbb{R}^{n+1}, d𝕊n(p,q)d_{\mathbb{S}^{n}}(p,q) is the angle between pp and qq.

The Gromov-Hausdorff (GH) distance provides a quantitative measure of how far two metric spaces are from being isometric. Since being introduced by Edwards ([Ed], 1975) and Gromov ([Gr1, Gr2], 1981), it has proved useful in the study of, for example, shapes formed by point cloud data [MS, BBK], convergence results for sequences of Riemannian manifolds [CC, PW, C1, C2], differentiability results in metric measure spaces [Ke, Ch] or the stability of topological invariants of metric spaces under small deformations (see [Pe] and the present article).

However, it is hard to find the exact value of the GH distance between two given metric spaces. There have been recent efforts to determine the precise value of the GH distance between spheres. In [LMS, Theorem B] it is proved that dGH(𝕊n,𝕊m)12arccos(1m+1)d_{\textup{GH}}(\mathbb{S}^{n},\mathbb{S}^{m})\geq\frac{1}{2}\arccos\left(\frac{-1}{m+1}\right) for all n>mn>m which implies that

dGH(𝕊n,𝕊1)π3d_{\textup{GH}}(\mathbb{S}^{n},\mathbb{S}^{1})\geq\frac{\pi}{3} (1)

for all n2n\geq 2. Via these lower bounds and matching upper bounds, the authors find the precise value of the GH distance for the pairs (𝕊1,𝕊2)(\mathbb{S}^{1},\mathbb{S}^{2}), (𝕊1,𝕊3)(\mathbb{S}^{1},\mathbb{S}^{3}), and (𝕊2,𝕊3)(\mathbb{S}^{2},\mathbb{S}^{3}). Similarly, in [Ka, Lemma 2.3] it was proved that for any interval II\subseteq\mathbb{R} we have dGH(I,𝕊1)π3d_{\textup{GH}}(I,\mathbb{S}^{1})\geq\frac{\pi}{3} (see also [LMS, Prop. B.1]; in [JT, Theorem 6.8] the exact GH distances from 𝕊1\mathbb{S}^{1} to intervals of any length λ[0,)\lambda\in[0,\infty) are computed). In [ABC] the authors identify lower bounds for dGH(𝕊m,𝕊n)d_{\textup{GH}}(\mathbb{S}^{m},\mathbb{S}^{n}) that are often tighter than the ones from [LMS].

Via considerations related to persistent homology and the filling radius, in [LMO, Remark 9.19] it was deduced that for any compact geodesic metric space XX one has dGH(X,𝕊1)π6d_{\textup{GH}}(X,\mathbb{S}^{1})\geq\frac{\pi}{6} and, as a generalization of Equation (1), the following conjecture was formulated:

Conjecture 1.1 ([LMO, Conjecture 4]).

For any compact, simply connected geodesic space XX we have dGH(X,𝕊1)π3d_{\textup{GH}}(X,\mathbb{S}^{1})\geq\frac{\pi}{3}.

In 1.1, the condition that XX is geodesic is necessary to relate the metric of XX to the fact that it is simply connected; if not, we could consider the space X=𝕊1{1}X=\mathbb{S}^{1}\setminus\{1\}, with its metric inherited from 𝕊1\mathbb{S}^{1}. Then XX is simply connected, but we clearly have dGH(X,𝕊1)=0d_{\textup{GH}}(X,\mathbb{S}^{1})=0. The present article gives a complete answer to 1.1; we state the main results after recalling some standard notation.

Let (X,dX)(X,d_{X}) be a metric space, and let A,BXA,B\subseteq X be nonempty and let xXx\in X. The distance from xx to AA is defined as dX(x,A)=infaAdX(x,a)d_{X}(x,A)=\inf_{a\in A}d_{X}(x,a), and the Hausdorff distance between AA and BB is given by

dHX(A,B)=max(supaAdX(a,B),supbBdX(b,A)).d^{X}_{\text{H}}(A,B)=\max\left(\sup_{a\in A}d_{X}(a,B),\sup_{b\in B}d_{X}(b,A)\right).

Given two nonempty metric spaces (X,dX)(X,d_{X}) and (Y,dY)(Y,d_{Y}), we write (X,dX)(Y,dY)(X,d_{X})\cong(Y,d_{Y}), or just XYX\cong Y when the metrics are clear, whenever (X,dX)(X,d_{X}) and (Y,dY)(Y,d_{Y}) are isometric. The Gromov-Hausdorff (GH) distance between XX and YY is defined as the value in [0,][0,\infty] given by

dGH(X,Y)=inf{dHZ(X,Y);(Z,dZ) metric space; X,YZ;XX;YY}.d_{\textup{GH}}(X,Y)=\inf\{d^{Z}_{\text{H}}(X^{\prime},Y^{\prime});(Z,d_{Z})\text{ metric space; }X^{\prime},Y^{\prime}\subseteq Z;X^{\prime}\cong X;Y^{\prime}\cong Y\}.

Recall that a metric space (X,dX)(X,d_{X}) is a length space when for all x,xXx,x^{\prime}\in X, the distance dX(x,x)d_{X}(x,x^{\prime}) is the infimum of lengths of paths γ:[0,1]X\gamma:[0,1]\to X with γ(0)=x\gamma(0)=x and γ(1)=x\gamma(1)=x^{\prime} (cf. [BBI, Chapter 2]). If in addition we have that for all x,xXx,x^{\prime}\in X there is a path γ\gamma from xx to xx^{\prime} with length dX(x,x)d_{X}(x,x^{\prime}), we say that XX is geodesic.

Theorem 1.2.

There exists a simply connected geodesic space E with dGH(E,𝕊1)=π4d_{\textup{GH}}(\textup{E},\mathbb{S}^{1})=\frac{\pi}{4}.

Theorem 1.3.

Let X,YX,Y be length spaces. Suppose there is a constant D>0D>0 such that dGH(X,Y)<Dd_{\textup{GH}}(X,Y)<D and all loops of diameter <4D<4D are nulhomotopic in XX and YY. Then π1(X)\pi_{1}(X) and π1(Y)\pi_{1}(Y) are isomorphic.

Corollary 1.4.

Any simply connected length space X satisfies dGH(X,𝕊1)π4d_{\textup{GH}}(X,\mathbb{S}^{1})\geq\frac{\pi}{4}.

To prove 1.4 from Theorem 1.3 note that every loop α:[0,1]𝕊1\alpha:[0,1]\to\mathbb{S}^{1} with diameter <π<\pi is nulhomotopic. So if there was a simply connected space XX with dGH(X,𝕊1)<π4d_{\textup{GH}}(X,\mathbb{S}^{1})<\frac{\pi}{4}, then we could choose the constant DD from Theorem 1.3 to be just below π4\frac{\pi}{4}, which would give a contradiction as π1(𝕊1)0\pi_{1}(\mathbb{S}^{1})\neq 0.

Section 2 is almost entirely devoted to the proof of Theorem 1.3. At the end of the section we also give a lower bound for the distances from simply connected length spaces to 𝕊1\mathbb{S}^{1} with the metric inherited from 2\mathbb{R}^{2}.

Remark 1.5.

As was well pointed out to me by F. Mémoli, the arguments used in Section 2 are similar to those in Petersen’s article [Pe]. The Theorem in [Pe, Section 4] claims that, under adequate conditions, if the GH distance between two metric spaces X,YX,Y is small enough, then they are homotopy equivalent; in our case, as we are comparing simply connected spaces with 𝕊1\mathbb{S}^{1}, it will be enough to give conditions on dGH(X,Y)d_{\textup{GH}}(X,Y) such that π1(X)π1(Y)\pi_{1}(X)\cong\pi_{1}(Y). This allows us to give better bounds for dGH(X,Y)d_{\textup{GH}}(X,Y) for our purposes; for example, in the Main Lemma of [Pe, Section 2], if one sets n=1n=1 and if XX is a length space instead of a LGC(ρ)0{}^{0}(\rho) space, then one can improve Petersen’s conclusion f¯(Δ)B(f(v),ρ1(ε))\overline{f}(\Delta)\subseteq B(f(v),\rho_{1}(\varepsilon)) (with ρ1(ε)2ε\rho_{1}(\varepsilon)\geq 2\varepsilon) to the sharper condition f¯(Δ)B(f(v),ε)\overline{f}(\Delta)\subseteq B(f(v),\varepsilon).

Section 3 is devoted to the construction of a simply connected geodesic space E with dGH(E,𝕊1)π4d_{\textup{GH}}(\mathrm{E},\mathbb{S}^{1})\leq\frac{\pi}{4}. This, in conjunction with 1.4, proves Theorem 1.2. In Section 3 we will not use the definition of Gromov-Hausdorff distance given above; we now recall (cf. §7.3 in [BBI]) an equivalent definition based on correspondences between sets, which will be more convenient for our purposes.

Given two sets XX and YY, we say a relation RX×YR\subseteq X\times Y is a correspondence between XX and YY if πX(R)=X\pi_{X}(R)=X and πY(R)=Y\pi_{Y}(R)=Y, where πX:X×YX\pi_{X}:X\times Y\to X and πY:X×YY\pi_{Y}:X\times Y\to Y are the coordinate projections. If (X,dX)(X,d_{X}) and (Y,dY)(Y,d_{Y}) are metric spaces, we define the distortion of a nonempty relation RX×YR\subseteq X\times Y as

dis(R):=sup{|dX(x,x)dY(y,y)|;(x,y),(x,y)R}[0,].\text{dis}(R):=\sup\{|d_{X}(x,x^{\prime})-d_{Y}(y,y^{\prime})|;(x,y),(x^{\prime},y^{\prime})\in R\}\in[0,\infty]. (2)

In [BBI, Theorem 7.3.25] it is proved that, if (X,dX),(Y,dY)(X,d_{X}),(Y,d_{Y}) are nonempty metric spaces, then

dGH(X,Y)=12inf{dis(R);RX×Y correspondence between X and Y}.d_{\textup{GH}}(X,Y)=\frac{1}{2}\inf\{\text{dis}(R);R\subseteq X\times Y\text{ correspondence between $X$ and }Y\}. (3)

In Section 4 we adress the question of whether the space E\mathrm{E} we construct to prove Theorem 1.2, which is a compact \mathbb{R}-tree, is the unique space satisfying Theorem 1.2. It turns out that it is not hard to use E to construct a big family of spaces that satisfy Theorem 1.2 (see 4.2), and that there are \mathbb{R}-trees which satisfy Theorem 1.2 and are not isometric to E (see Note 4.6). However, in 4.5 we prove a minimality result for our space E: it is, up to isometry, the only complete \mathbb{R}-tree of length at most 5π4\frac{5\pi}{4} which satisfies Theorem 1.2.

Acknowledgements. Special thanks to Facundo Mémoli for introducing the author to the topic of Gromov-Hausdorff distances and providing guidance while writing this article. The author gratefully acknowledges support from the grants BSF 2020124 and NSF CCF AF 2310412.

2 Proof of Theorem 1.3

Lemma 2.1.

Let (Y,dY)(Y,d_{Y}) be a length space such that, for some constant C>0C>0, all loops of diameter <C<C are contractible in YY, and let y0Yy_{0}\in Y. Then any two loops β,β:[0,1]Y\beta,\beta^{\prime}:[0,1]\to Y based at y0y_{0} satisfying dY(β(t),β(t))<Cd_{Y}(\beta(t),\beta^{\prime}(t))<C for all t[0,1]t\in[0,1] are path-homotopic.

Proof.

Let δ>0\delta>0 be such that dY(β(t),β(t))<C2δd_{Y}(\beta(t),\beta^{\prime}(t))<C-2\delta for all t[0,1]t\in[0,1].

Let NN\in\mathbb{N} be big enough that for all n=1,,Nn=1,\dots,N, the paths βn:=β|[n1N,nN]\beta_{n}:=\beta|_{\left[\frac{n-1}{N},\frac{n}{N}\right]} and βn:=β|[n1N,nN]\beta^{\prime}_{n}:=\beta^{\prime}|_{\left[\frac{n-1}{N},\frac{n}{N}\right]} have diameter <δ<\delta. Note that β=β1βN\beta=\beta_{1}\cdots\beta_{N} and β=β1βN\beta^{\prime}=\beta^{\prime}_{1}\cdots\beta^{\prime}_{N}. Also, for each n=1,,N1n=1,\dots,N-1 let γn:[0,1]Y\gamma_{n}:[0,1]\to Y be a path with γn(0)=β(nN),γn(1)=β(nN)\gamma_{n}(0)=\beta\left(\frac{n}{N}\right),\gamma_{n}(1)=\beta^{\prime}\left(\frac{n}{N}\right) and with length <C2δ<C-2\delta, as in Figure 1.

Refer to caption
Fig. 1: The loops β\beta and β\beta^{\prime}, with paths γ1,,γN\gamma_{1},\dots,\gamma_{N} between them. Note that the 22 triangles and the N2N-2 rectangles in the figure are all nulhomotopic, as they have diameter <C<C.

Now consider the sequence of loops β=p0,p1,,pN=β\beta=p_{0},p_{1},\dots,p_{N}=\beta^{\prime}, where for n=1,,N1n=1,\dots,N-1 we have pn=β1βnγnβn+1βNp_{n}=\beta_{1}\dots\beta_{n}\gamma_{n}\beta^{\prime}_{n+1}\cdots\beta^{\prime}_{N}. Then for all n=1,,Nn=1,\dots,N, pn1p_{n-1} is homotopic to pnp_{n}. Indeed, for n=2,,N1n=2,\dots,N-1 (the cases n=1,Nn=1,N are easier) we have

pnpn11=(β1βn1)βnγn(βn)1(γn1)1(β1βn1)1,p_{n}p_{n-1}^{-1}=(\beta_{1}\cdots\beta_{n-1})\beta_{n}\gamma_{n}(\beta^{\prime}_{n})^{-1}(\gamma_{n-1})^{-1}(\beta_{1}\cdots\beta_{n-1})^{-1},

and the loop βnγn(βn)1(γn1)1\beta_{n}\gamma_{n}(\beta^{\prime}_{n})^{-1}(\gamma_{n-1})^{-1} is null because it has diameter <C<C (this follows from the facts that βn,βn1\beta_{n},\beta_{n-1} have diameter <δ<\delta and γn,γn1\gamma_{n},\gamma_{n-1} have length <C2δ<C-2\delta). We conclude then that β\beta and β\beta^{\prime} are homotopic, as we wanted. ∎

Theorem 2.2.

Let (X,dX),(Y,dY)(X,d_{X}),(Y,d_{Y}) be length spaces and let D>dGH(X,Y)D>d_{\textup{GH}}(X,Y). Suppose XX and YY are embedded isometrically into a metric space (Z,dZ)(Z,d_{Z}) with dHZ(X,Y)<Dd_{\mathrm{H}}^{Z}(X,Y)<D. Let x0X,y0Yx_{0}\in X,y_{0}\in Y with dZ(x0,y0)<Dd_{Z}(x_{0},y_{0})<D. Then for any loop α:[0,1]X\alpha:[0,1]\to X based at x0x_{0} there is a loop β:[0,1]Y\beta:[0,1]\to Y based at y0y_{0} such that dZ(α(t),β(t))<2Dd_{Z}(\alpha(t),\beta(t))<2D for all t[0,1]t\in[0,1].

Moreover, if all loops (not necessarily based at y0y_{0}) of diameter <4D<4D in YY are contractible in YY, then the function αβ\alpha\mapsto\beta defined above gives a well defined, surjective111If XX is not a length space, then Theorem 2.2 still holds with the same proof, except that the map Φ\Phi need not be surjective. map Φ:π1(X,x0)π1(Y,y0)\Phi:\pi_{1}(X,x_{0})\to\pi_{1}(Y,y_{0}).

Proof.

Let δ>0\delta>0 be such that dHZ(X,Y)<Dδd_{\text{H}}^{Z}(X,Y)<D-\delta. Let NN\in\mathbb{N} be big enough that diam(α|[n1N,nN])<δ\left(\alpha|_{\left[\frac{n-1}{N},\frac{n}{N}\right]}\right)<\delta for all n=1,,Nn=1,\dots,N, and for each n=0,,Nn=0,\dots,N define β(nN)\beta\left(\frac{n}{N}\right) to be a point of YY such that dZ(α(nN),β(nN))<Dδd_{Z}\left(\alpha\left(\frac{n}{N}\right),\beta\left(\frac{n}{N}\right)\right)<D-\delta (one can choose β(0)=β(1)=y0\beta(0)=\beta(1)=y_{0}). By the triangular inequality, dY(β(n1N),β(nN))<2(Dδ)+δ<2Dd_{Y}\left(\beta\left(\frac{n-1}{N}\right),\beta\left(\frac{n}{N}\right)\right)<2(D-\delta)+\delta<2D, thus we can define β|[n1N,nN]\beta|_{\left[\frac{n-1}{N},\frac{n}{N}\right]} to be a path from β(n1N)\beta\left(\frac{n-1}{N}\right) to β(nN)\beta\left(\frac{n}{N}\right) of length <2D<2D.

Note that all points in β|[n1N,nN]\beta|_{\left[\frac{n-1}{N},\frac{n}{N}\right]} are at distance <D<D of either β(n1N)\beta\left(\frac{n-1}{N}\right) or β(nN)\beta\left(\frac{n}{N}\right). Moreover, β(n1N)\beta\left(\frac{n-1}{N}\right) and β(nN)\beta\left(\frac{n}{N}\right) are both at distance <D<D from all points in α|[n1N,nN]\alpha|_{\left[\frac{n-1}{N},\frac{n}{N}\right]}. Thus, all points in β|[n1N,nN]\beta|_{\left[\frac{n-1}{N},\frac{n}{N}\right]} are at distance <D+(Dδ)<D+(D-\delta) from all points in α|[n1N,nN]\alpha|_{\left[\frac{n-1}{N},\frac{n}{N}\right]}, and in particular dZ(β(t),α(t))<2Dd_{Z}(\beta(t),\alpha(t))<2D for all t[0,1]t\in[0,1], as we wanted. This concludes the proof of the first part of the theorem.

Suppose now that all loops of diameter <4D<4D are contractible in YY. Suppose that α,α:[0,1]X\alpha,\alpha^{\prime}:[0,1]\to X are path-homotopic loops (the path-homotopy being inside XX) based at x0x_{0}, and let β,β:[0,1]Y\beta,\beta^{\prime}:[0,1]\to Y be loops based at y0y_{0} such that dZ(α(t),β(t))<2Dd_{Z}(\alpha(t),\beta(t))<2D and dZ(α(t),β(t))<2Dd_{Z}(\alpha^{\prime}(t),\beta^{\prime}(t))<2D for all t[0,1]t\in[0,1]. We want to prove that β,β\beta,\beta^{\prime} are homotopic.

To do it, let δ>0\delta>0 be such that dHZ(X,Y)<Dδd_{\text{H}}^{Z}(X,Y)<D-\delta and such that dZ(α(t),β(t))<2Dδd_{Z}(\alpha(t),\beta(t))<2D-\delta and dZ(α(t),β(t))<2Dδd_{Z}(\alpha^{\prime}(t),\beta^{\prime}(t))<2D-\delta for all t[0,1]t\in[0,1]. Let (αt)t[0,1](\alpha_{t})_{t\in[0,1]} be a path-homotopy in XX from α0=α\alpha_{0}=\alpha to α1=α\alpha_{1}=\alpha^{\prime}. By uniform continuity of the homotopy, we can choose big enough NN such that for all n=1,,Nn=1,\dots,N and for all t[0,1]t\in[0,1] we have dX(αn1N(t),αnN(t))<δd_{X}\left(\alpha_{\frac{n-1}{N}}(t),\alpha_{\frac{n}{N}}(t)\right)<\delta. Now by the first part of the theorem, we can let βnN\beta_{\frac{n}{N}} be a loop in YY based at y0y_{0} such that dZ(αnN(t),βnN(t))<2(Dδ)d_{Z}(\alpha_{\frac{n}{N}}(t),\beta_{\frac{n}{N}}(t))<2(D-\delta) for all t[0,1]t\in[0,1] (with β0=β\beta_{0}=\beta and β1=β\beta_{1}=\beta^{\prime}). By the triangular inequality we have for all n=1,,Nn=1,\dots,N and t[0,1]t\in[0,1] that dZ(βn1N(t),βnN(t))4(Dδ)+δ<4Dd_{Z}\left(\beta_{\frac{n-1}{N}}(t),\beta_{\frac{n}{N}}(t)\right)\leq 4(D-\delta)+\delta<4D, thus by Lemma 2.1, βn1N\beta_{\frac{n-1}{N}} and βnN\beta_{\frac{n}{N}} are path-homotopic. We conclude then that β0=β\beta_{0}=\beta and β1=β\beta_{1}=\beta^{\prime} are path-homotopic. This proves that the map Φ:π1(X,x0)π1(Y,y0)\Phi:\pi_{1}(X,x_{0})\to\pi_{1}(Y,y_{0}) is well defined.

To check that Φ\Phi is surjective note that, by the first part of the theorem, for any loop β:[0,1]Y\beta:[0,1]\to Y based at y0y_{0} we can find a loop α:[0,1]X\alpha:[0,1]\to X based at x0x_{0} such that dZ(β(t),α(t))<2Dd_{Z}(\beta(t),\alpha(t))<2D for all tt, so Φ([α])=[β]\Phi([\alpha])=[\beta]. ∎

Proof of Theorem 1.3.

Let (Z,dZ)(Z,d_{Z}) contain X,YX,Y isometrically and choose basepoints x0,y0x_{0},y_{0} in X,YX,Y respectively with dZ(x0,y0)<Dd_{Z}(x_{0},y_{0})<D. Define maps Φ:π1(X,x0)π1(Y,y0)\Phi:\pi_{1}(X,x_{0})\to\pi_{1}(Y,y_{0}) and Ψ:π1(Y,y0)π1(X,x0)\Psi:\pi_{1}(Y,y_{0})\to\pi_{1}(X,x_{0}) as in Theorem 2.2. Then the composition ΨΦ\Psi\circ\Phi is the identity (similarly with ΦΨ\Phi\circ\Psi): indeed, for any loop α:[0,1]X\alpha:[0,1]\to X based at x0x_{0}, Φ([α])\Phi([\alpha]) will be [β][\beta] for some loop β:[0,1]Y\beta:[0,1]\to Y based at y0y_{0} such that dZ(α(t),β(t))<2Dd_{Z}(\alpha(t),\beta(t))<2D for all t[0,1]t\in[0,1]. But then we also have Ψ([β])=[α]\Psi([\beta])=[\alpha], so ΨΦ([α])=[α]\Psi\circ\Phi([\alpha])=[\alpha]. ∎

Note 2.3.

One can use the ideas from this section to find lower bounds for the distance from any simply connected length space to 𝕊1\mathbb{S}^{1} with the Euclidean metric d2d_{\mathbb{R}^{2}} inherited from 2\mathbb{R}^{2}. In (𝕊1,d2)(\mathbb{S}^{1},d_{\mathbb{R}^{2}}), the conclusion of Lemma 2.1 holds for C=2C=2, that is, any two loops β0,β1:[0,1]𝕊1\beta_{0},\beta_{1}:[0,1]\to\mathbb{S}^{1} with d2(β0(t),β1(t))<2d_{\mathbb{R}^{2}}(\beta_{0}(t),\beta_{1}(t))<2 for all tt are path-homotopic, with a homotopy (βs)s[0,1](\beta_{s})_{s\in[0,1]} between them defined by, for each tt, letting sβs(t)s\mapsto\beta_{s}(t) be the shortest geodesic from β0(t)\beta_{0}(t) to β1(t)\beta_{1}(t). Also note that for any D>0D>0 and any two points x,y𝕊1x,y\in\mathbb{S}^{1} with d2(x,y)2Dd_{\mathbb{R}^{2}}(x,y)\leq 2D, the points in the shortest path from xx to yy will not be at distance >221D2>\sqrt{2-2\sqrt{1-D^{2}}} from the set {x,y}\{x,y\}. That and the proof of Theorem 2.2 imply that, if we have a length space XX at Hausdorff distance <D<D from 𝕊1\mathbb{S}^{1} inside some bigger metric space (Z,dZ)(Z,d_{Z}), then for each loop α:[0,1]X\alpha:[0,1]\to X we can find a loop β:[0,1]𝕊1\beta:[0,1]\to\mathbb{S}^{1} such that for all tt, dZ(α(t),β(t))<D+221D2d_{Z}(\alpha(t),\beta(t))<D+\sqrt{2-2\sqrt{1-D^{2}}}. If 2(D+221D2)<C=22\left(D+\sqrt{2-2\sqrt{1-D^{2}}}\right)<C=2, as in Theorem 2.2 we obtain a well defined surjective map π1(X)π1(𝕊1)\pi_{1}(X)\to\pi_{1}\left(\mathbb{S}^{1}\right), so XX cannot be simply connected.

This proves that the GH distance DD from (𝕊1,d2)(\mathbb{S}^{1},d_{\mathbb{R}^{2}}) to a simply connected length space cannot satisfy D+221D2<1D+\sqrt{2-2\sqrt{1-D^{2}}}<1. Thus we obtain a lower bound for DD of approximately 0.491650.49165. We could not construct an optimal example similar to the one in Section 3 and we currently see no reason why this lower bound should be optimal, which naturally leads to the following question:

Question 2.4.

What is the infimal distance from a simply connected geodesic space to (𝕊1,d2)(\mathbb{S}^{1},d_{\mathbb{R}^{2}})?

3 A simply connected geodesic space E\mathrm{E} with
d(E,𝕊1)GH=π4{}_{\textup{GH}}(\mathrm{E},\mathbb{S}^{1})=\frac{\pi}{4}

Figure 2 shows two metric graphs (see [BBI, Definition 3.2.9.] for a definition of metric graph) which we will call E and 𝕊1\mathbb{S}^{1} for obvious reasons, with all edges having length π4\frac{\pi}{4}. The graph E has five edges, a,b,c,da,b,c,d and ee. We have also labelled the eight edges of 𝕊1\mathbb{S}^{1} with the letters a,b,c,d,ea,b,c,d,e.

We define a function Φ:𝕊1E\Phi:\mathbb{S}^{1}\to\textup{E} which sends each of the 88 edges of 𝕊1\mathbb{S}^{1} isometrically to one of the 55 segments a,b,c,d,ea,b,c,d,e of E as indicated by the labels and orientations in the figure. This specification of Φ\Phi gives two possible values P,QEP,Q\in\textup{E} to Φ(1)\Phi(1); for definiteness we choose Φ(1)=P\Phi(1)=P. Note that Φ\Phi is locally 11-Lipschitz except at 11, where there is a jump of length π2\frac{\pi}{2} from QQ to PP.

Refer to caption
Fig. 2: The spaces E and 𝕊1\mathbb{S}^{1}.
Claim 3.1.

The graph of Φ\Phi, R:={(x,Φ(x));x𝕊1}𝕊1×ER:=\{(x,\Phi(x));x\in\mathbb{S}^{1}\}\subseteq\mathbb{S}^{1}\times\textup{E}, is a correspondence between 𝕊1\mathbb{S}^{1} and E with dis(R)π2\text{dis}(R)\leq\frac{\pi}{2}.

Note that, by Equation 3, the claim above implies that dGH(E,𝕊1)π4d_{\textup{GH}}(\textup{E},\mathbb{S}^{1})\leq\frac{\pi}{4}. Since E is a simply connected geodesic space, by 1.4 we obtain dGH(E,𝕊1)=π4d_{\textup{GH}}(\textup{E},\mathbb{S}^{1})=\frac{\pi}{4}, which is the goal of this section.

Proof of 3.1.

First note that Φ\Phi is surjective, so Φ\Phi is indeed a correspondence between 𝕊1\mathbb{S}^{1} and EE. It remains to prove that for all x,x𝕊1x,x^{\prime}\in\mathbb{S}^{1} we have

|dE(Φ(x),Φ(x))d𝕊1(x,x)|π2.\big{|}d_{\textup{E}}(\Phi(x),\Phi(x^{\prime}))-d_{\mathbb{S}^{1}}(x,x^{\prime})\big{|}\leq\frac{\pi}{2}.

First we prove the inequality dE(Φ(x),Φ(x))d𝕊1(x,x)π2d_{\textup{E}}(\Phi(x),\Phi(x^{\prime}))-d_{\mathbb{S}^{1}}(x,x^{\prime})\leq\frac{\pi}{2}. To do this, we may assume xxx\neq x^{\prime} and consider the geodesic segment [x,x][x,x^{\prime}], given by a unit speed curve γ:[0,d]𝕊1\gamma:[0,d]\to\mathbb{S}^{1} with γ(0)=x\gamma(0)=x, γ(1)=x\gamma(1)=x^{\prime} and d:=d𝕊1(x,x)π2d:=d_{\mathbb{S}^{1}}(x,x^{\prime})\leq\frac{\pi}{2}. Then the curve Φγ:[0,d]E\Phi\circ\gamma:[0,d]\to\textup{E} either is 11-Lipschitz, in which case dE(Φ(x),Φ(x))len(γ)dd_{\textup{E}}(\Phi(x),\Phi(x^{\prime}))\leq\text{len}(\gamma)\leq d, or it has a jump of length π2\frac{\pi}{2} at some point s[0,d]s\in[0,d]. In this case, Φγ\Phi\circ\gamma is still 11-Lipschitz in [0,s)[0,s) and (s,d](s,d], so we have the inequality we wanted:

dE(Φ(x),Φ(x))length(γ)s+π2+(sd)=π2+d=π2+d𝕊1(x,x).d_{\textup{E}}(\Phi(x),\Phi(x^{\prime}))\leq\text{length}(\gamma)\leq s+\frac{\pi}{2}+(s-d)=\frac{\pi}{2}+d=\frac{\pi}{2}+d_{\mathbb{S}^{1}}(x,x^{\prime}).

Now we prove the remaining inequality, d𝕊1(x,x)dE(Φ(x),Φ(x))π2d_{\mathbb{S}^{1}}(x,x^{\prime})-d_{\textup{E}}(\Phi(x),\Phi(x^{\prime}))\leq\frac{\pi}{2}. When x,xx,x^{\prime} are antipodal, we have equality: dE(Φ(x),Φ(x))=π2d_{\textup{E}}(\Phi(x),\Phi(-x))=\frac{\pi}{2}. This can be checked using the definition of Φ\Phi in the eight edges of 𝕊1\mathbb{S}^{1}.

For general points xxx\neq x^{\prime}, note that 11 cannot be in both of the geodesic segments [x,x][-x^{\prime},x] and [x,x][x^{\prime},-x], because they are antipodal segments of length <π<\pi. Suppose without loss of generality that 11 is not in [x,x][-x^{\prime},x], so that, if γ:[0,d]𝕊1\gamma:[0,d]\to\mathbb{S}^{1} is the unit speed geodesic with γ(0)=x\gamma(0)=x^{\prime}, γ(d)=x\gamma(d)=-x and d=d𝕊1(x,x)d=d_{\mathbb{S}^{1}}(x^{\prime},-x), then the path Φγ\Phi\circ\gamma is 11-Lipschitz. Now consider the function

h:[0,d];h(t):=d𝕊1(x,γ(t))dE(Φ(x),Φ(γ(t))).h:[0,d]\to\mathbb{R};h(t):=d_{\mathbb{S}^{1}}(x,\gamma(t))-d_{\textup{E}}(\Phi(x),\Phi(\gamma(t))).

Then hh is increasing, because ddtd𝕊1(x,γ(t))=1\frac{d}{dt}d_{\mathbb{S}^{1}}(x,\gamma(t))=1 and the map tdE(Φ(x),Φ(γ(t)))t\mapsto d_{\textup{E}}(\Phi(x),\Phi(\gamma(t))) is 11-Lipschitz. Thus, as we wanted,

π2=d𝕊1(x,x)dE(Φ(x),Φ(x))=Φ(d)Φ(0)=d𝕊1(x,x)dE(Φ(x),Φ(x)).\frac{\pi}{2}=d_{\mathbb{S}^{1}}(x,-x)-d_{\textup{E}}(\Phi(x),\Phi(-x))=\Phi(d)\geq\Phi(0)=d_{\mathbb{S}^{1}}(x,x^{\prime})-d_{\textup{E}}(\Phi(x),\Phi(x^{\prime})).

4 A uniqueness result for the space E\mathrm{E}

It is natural to ask whether E\mathrm{E} is, up to isometry, the only geodesic simply connected space with GH distance π4\frac{\pi}{4} to 𝕊1\mathbb{S}^{1}. We now use E\mathrm{E} to obtain a big family of simply connected geodesic spaces which are exactly at GH distance π4\frac{\pi}{4} from 𝕊1\mathbb{S}^{1}.

Proposition 4.1.

Let (X,dX)(X,d_{X}), (Y,dY)(Y,d_{Y}) and (Z,dZ)(Z,d_{Z}) be nonempty metric spaces, and suppose diam(Z)2dGH(X,Y)\text{diam}(Z)\leq 2d_{\textup{GH}}(X,Y). Then the space X×ZX\times Z, with the metric

dX×Z((x,z),(x,z)):=max(dX(x,x),dZ(z,z)),d_{X\times Z}((x,z),(x^{\prime},z^{\prime})):=\max(d_{X}(x,x^{\prime}),d_{Z}(z,z^{\prime})),

satisfies dGH(X×Z,Y)dGH(X,Y)d_{\mathrm{GH}}(X\times Z,Y)\leq d_{\textup{GH}}(X,Y).

Proof.

For any correspondence RX×YR\subseteq X\times Y, which by Equation 3 satisfies dis(R)2dGH(X,Y)diam(Z)(R)\geq 2d_{\textup{GH}}(X,Y)\geq\text{diam}(Z), we can define the correspondence

RZ:={((x,z),y);(x,y)R,zZ}(X×Z)×Y.R_{Z}:=\{((x,z),y);(x,y)\in R,z\in Z\}\subseteq(X\times Z)\times Y.

We will be done by Equation 3 if we prove that dis(RZ)dis(R)\text{dis}(R_{Z})\leq\text{dis}(R). Indeed, letting ((x,z),y),((x,z),y)RZ((x,z),y),((x^{\prime},z^{\prime}),y^{\prime})\in R_{Z}, we have

dY(y,y)dX×Z((x,z),(x,z))\displaystyle d_{Y}(y,y^{\prime})-d_{X\times Z}((x,z),(x^{\prime},z^{\prime})) dY(y,y)dX(x,x)\displaystyle\leq d_{Y}(y,y^{\prime})-d_{X}(x,x^{\prime})
dis(R).\displaystyle\leq\text{dis}(R).
dX×Z((x,z),(x,z))dY(y,y)\displaystyle d_{X\times Z}((x,z),(x^{\prime},z^{\prime}))-d_{Y}(y,y^{\prime}) =max(dX(x,x)dY(y,y),dZ(z,z)dY(y,y))\displaystyle=\max(d_{X}(x,x^{\prime})-d_{Y}(y,y^{\prime}),d_{Z}(z,z^{\prime})-d_{Y}(y,y^{\prime}))
max(dis(R),diam(Z))\displaystyle\leq\max(\text{dis}(R),\text{diam}(Z))
dis(R).\displaystyle\leq\text{dis}(R).\qed
Corollary 4.2.

Let (X,dX)(X,d_{X}) be a nonempty simply connected geodesic space with diam(X)π2\text{diam}(X)\leq\frac{\pi}{2}. Give X×EX\times\mathrm{E} the metric

dX×E((x,e),(x,e))=max(dX(x,x),dE(e,e)).d_{X\times\mathrm{E}}((x,e),(x^{\prime},e^{\prime}))=\max(d_{X}(x,x^{\prime}),d_{\mathrm{E}}(e,e^{\prime})).

Then (X×E,dX×E)(X\times\mathrm{E},d_{X\times\mathrm{E}}) is a simply connected geodesic space and dGH(X×E)=π4d_{\textup{GH}}(X\times\mathrm{E})=\frac{\pi}{4}.

Proof.

The space X×EX\times\mathrm{E} is simply connected and geodesic (see [, Page 2]). It then follows from 4.1 and 1.4 that dGH(X×E,𝕊1)=π4d_{\textup{GH}}(X\times\mathrm{E},\mathbb{S}^{1})=\frac{\pi}{4}. ∎

Note 4.3.

More generally, the proof of 4.2 shows that any geodesic, simply connected subspace ZX×EZ\subseteq X\times\mathrm{E} such that the coordinate projection πE:ZE\pi_{\mathrm{E}}:Z\to\mathrm{E} is surjective will also be at distance π4\frac{\pi}{4} from 𝕊1\mathbb{S}^{1}.

So E is not the only geodesic, simply connected space at distance π4\frac{\pi}{4} from 𝕊1\mathbb{S}^{1}. However, we prove below that E is the only complete tree with minimal length at distance π4\frac{\pi}{4} from 𝕊1\mathbb{S}^{1}; to formalize this statement we will need to introduce some concepts and notation:

Definition 4.4 ([Be, Defs. 2.1, 2.2.]).

An arc in a metric space XX is a subspace of XX homeomorphic to the interval [0,1][0,1]. A metric space (X,dX)(X,d_{X}) is an \mathbb{R}-tree if for every x,xXx,x^{\prime}\in X there is a unique arc with endpoints x,yx,y and this arc is a geodesic segment (i.e., isometric to an interval of \mathbb{R}). We denote by [x,x]X[x,x^{\prime}]\subseteq X the geodesic segment from xx to xx^{\prime}. We will define the length of the \mathbb{R}-tree XX to be the supremum of all sums of lengths of disjoint arcs contained in XX.

Note that all \mathbb{R}-trees are geodesic by definition, and the distance between any two points x,xx,x^{\prime} of an \mathbb{R}-tree is the length of the arc [x,x][x,x^{\prime}]. Moreover, any \mathbb{R}-tree XX is contractible: indeed, for any x0Xx_{0}\in X we can define a homotopy (ft)t[0,1](f_{t})_{t\in[0,1]}, ft:XXf_{t}:X\to X, from a constant map to IdX by letting ft(x)f_{t}(x) be the point in [x,x0][x,x_{0}] at distance tdX(x,x0)t\cdot d_{X}(x,x_{0}) from x0x_{0}. Thus, by 1.4, any \mathbb{R}-tree XX satisfies dGH(X,𝕊1)π4d_{\mathrm{GH}}(X,\mathbb{S}^{1})\geq\frac{\pi}{4}.

Proposition 4.5.

Let (X,dX)(X,d_{X}) be a complete \mathbb{R}-tree with length 5π4\leq\frac{5\pi}{4} and such that dGH(X,𝕊1)=π4d_{\textup{GH}}(X,\mathbb{S}^{1})=\frac{\pi}{4}. Then (X,dX)(X,d_{X}) is isometric to E.

Note 4.6.

4.5 does not hold for finite simplicial graphs instead of \mathbb{R}-trees: for example, the metric space (𝕊1,d𝕊12)\left(\mathbb{S}^{1},\frac{d_{\mathbb{S}^{1}}}{2}\right) has length π\pi and is at GH distance π4\frac{\pi}{4} from 𝕊1\mathbb{S}^{1} (the identity map 𝕊1𝕊1\mathbb{S}^{1}\to\mathbb{S}^{1} gives an optimal correspondence).

The condition of having length 5π4\leq\frac{5\pi}{4} is also necessary to prove uniqueness of E in 4.5, that is, there are \mathbb{R}-trees XX not isometric to E and such that dGH(X,𝕊1)=π4d_{\textup{GH}}(X,\mathbb{S}^{1})=\frac{\pi}{4}222This fact doesn’t seem to follow from Note 4.3, so we give a separate construction.. Indeed, consider a metric tree E’ obtained from E by increasing the length of the edge cc (see Figure 2) from π4\frac{\pi}{4} to π2\frac{\pi}{2}. We can consider E as a subspace of E’, with s:=E’Es:=\textup{E'}\setminus\textup{E} being a segment of length π4\frac{\pi}{4}. Letting R𝕊1×ER\subseteq\mathbb{S}^{1}\times\textup{E} be the correspondence from 3.1, we can define a correspondence

R=R({1}×s)𝕊1×E’,R^{\prime}=R\bigcup(\{-1\}\times s)\subseteq\mathbb{S}^{1}\times\textup{E'},

and it is not hard to check that dis(R)=π2\text{dis}(R^{\prime})=\frac{\pi}{2}.

Note 4.7.

Note that the condition ‘dGH(X,𝕊1)=π4d_{\textup{GH}}(X,\mathbb{S}^{1})=\frac{\pi}{4}’ can be equivalently expressed in the following way333To check that both statements are equivalent, one needs to use that \mathbb{R}-trees of finite length are compact, see 4.8.: there exist isometric copies of XX and 𝕊1\mathbb{S}^{1} inside a bigger metric space (Z,dZ)(Z,d_{Z}) such that the two following conditions are satisfied:

  1. 1.

    dZ(x,𝕊1)π4d_{Z}(x,\mathbb{S}^{1})\leq\frac{\pi}{4} for all xXx\in X.

  2. 2.

    dZ(p,X)π4d_{Z}(p,X)\leq\frac{\pi}{4} for all p𝕊1p\in\mathbb{S}^{1}.

Thus, 4.5 may be informally expressed as ‘besides E\mathrm{E}, no other complete \mathbb{R}-tree XX of length 5π4\leq\frac{5\pi}{4} can satisfy conditions 1 and 2 simultaneously’. If we remove condition 2, it is clear that many complete \mathbb{R}-trees of length 5π4\leq\frac{5\pi}{4} can satisfy condition 1 (e.g. the \mathbb{R}-tree formed by a single point). So we certainly need to use condition 2 in order to prove our theorem. However, one may at first think that condition 1 is not necessary to prove our uniqueness theorem, that is, that any \mathbb{R}-tree XX of length 5π4\leq\frac{5\pi}{4} satisfying condition 2 is isometric to E\mathrm{E}. That is, ‘if an \mathbb{R}-tree XX is smaller than EE, then we cannot place XX and 𝕊1\mathbb{S}^{1} in such a way that all points of 𝕊1\mathbb{S}^{1} are at distance π4\leq\frac{\pi}{4} of XX’. This is, however, not true. Indeed, consider the \mathbb{R}-tree XX shown below, which has four edges of length π4\frac{\pi}{4} and four extreme points p1,p2,p3,p4p_{1},p_{2},p_{3},p_{4}. Then XX has length π\pi, but if we divide 𝕊1\mathbb{S}^{1} into four intervals I1,I2,I3,I4I_{1},I_{2},I_{3},I_{4} of length π2\frac{\pi}{2}, one can give the disjoint union 𝕊1X\mathbb{S}^{1}\sqcup X a metric in which, for each i=1,2,3,4i=1,2,3,4, all the points of IiI_{i} are at distance π4\frac{\pi}{4} from pip_{i}.

[Uncaptioned image]
[Uncaptioned image]

The proof of 4.5 comprises the rest of this section. Fix a complete \mathbb{R}-tree (X,dX)(X,d_{X}) of length at most 5π4\frac{5\pi}{4} satisfying dGH(X,𝕊1)=π4d_{\mathrm{GH}}(X,\mathbb{S}^{1})=\frac{\pi}{4}.

Claim 4.8.

XX is compact.

Proof.

It is enough to prove that XX is totally bounded. If not, there is some ε>0\varepsilon>0 and points (xn)nX(x_{n})_{n\in\mathbb{N}}\in X with d(xi,xj)>4εd(x_{i},x_{j})>4\varepsilon if iji\neq j. So the balls Bn:=BX(xn,2ε)B_{n}:=B_{X}(x_{n},2\varepsilon) are pairwise disjoint, and each contains a segment of length ε\varepsilon, contradicting the fact that XX has finite length. ∎

As XX and 𝕊1\mathbb{S}^{1} are compact and dGH(X,𝕊1)=π4d_{\mathrm{GH}}(X,\mathbb{S}^{1})=\frac{\pi}{4}, it is proved in [CM, Proposition 1.1.] that there is a correspondence RX×𝕊1R\subseteq X\times\mathbb{S}^{1} achieving the minimal distortion dis(R)=π2\textup{dis}(R)=\frac{\pi}{2}, and such that RR is a closed subspace of X×𝕊1X\times\mathbb{S}^{1}.

So from now on, we fix a closed correspondence RX×𝕊1R\subseteq X\times\mathbb{S}^{1} with distortion π2\frac{\pi}{2}. For any xXx\in X we let R[x]:={p𝕊1;(x,p)R}R[x]:=\{p\in\mathbb{S}^{1};(x,p)\in R\}, and for any p𝕊1p\in\mathbb{S}^{1} we let R[p]:={xX;(x,p)R}R[p]:=\{x\in X;(x,p)\in R\} and we denote by p-p the antipodal of pp. We will fix two points x,x+Xx_{-},x_{+}\in X with dX(x,x+)=diam(X)d_{X}(x_{-},x_{+})=\text{diam}(X) and let x0x_{0} be the midpoint of [x,x+][x_{-},x_{+}]. Finally, we define the ‘projection’ map

Π:X[x,x+]\Pi:X\to[x_{-},x_{+}]

by the equation dX(x,Π(x))=dX(x,[x,x+])d_{X}(x,\Pi(x))=d_{X}(x,[x_{-},x_{+}]). Note that if two points x,xXx,x^{\prime}\in X have Π(x)Π(x)\Pi(x)\neq\Pi(x^{\prime}), then the union of [x,Π(x)],[Π(x),Π(x)][x,\Pi(x)],[\Pi(x),\Pi(x^{\prime})] and [Π(x),x][\Pi(x^{\prime}),x^{\prime}] is the geodesic arc from xx to xx^{\prime} (see figure Figure 3), so we have

dX(x,x)=dX(x,Π(x))+dX(Π(x),Π(x))+dX(Π(x),x).d_{X}(x,x^{\prime})=d_{X}(x,\Pi(x))+d_{X}(\Pi(x),\Pi(x^{\prime}))+d_{X}(\Pi(x^{\prime}),x^{\prime}). (4)
Refer to caption
Fig. 3: Geodesics between points x,xXx,x^{\prime}\in X with Π(x)Π(x)\Pi(x)\neq\Pi(x^{\prime}).

We will repeatedly use Equation 4 to compute distances in XX.

Definition 4.9.

A tripod is an \mathbb{R}-tree YY with three points y0,y1,y2Yy_{0},y_{1},y_{2}\in Y (the extreme points of YY) such that Y=[y0,y1][y0,y2][y1,y2]Y=[y_{0},y_{1}]\cup[y_{0},y_{2}]\cup[y_{1},y_{2}].

Claim 4.10.

We have dX(x,x+)=diam(X)=πd_{X}(x_{-},x_{+})=\text{diam}(X)=\pi, and the space XX is a tripod with extreme points x,x+x_{-},x_{+}, and a point x1x_{1} with dX(x1,Π(x1))=π4d_{X}(x_{1},\Pi(x_{1}))=\frac{\pi}{4}.

Proof.

We first deduce that diam(X)π\text{diam}(X)\geq\pi. To do it, note that any point xXx\in X is at distance diam(X)2\leq\frac{\text{diam}(X)}{2} from x0x_{0}: if not, depending on whether Π(x)[x,x0]\Pi(x)\in[x_{-},x_{0}] or Π(x)[x0,x+]\Pi(x)\in[x_{0},x_{+}] we would have dX(x,x+)>diam(X)d_{X}(x,x_{+})>\text{diam}(X) or dX(x,x)>diam(X)d_{X}(x,x_{-})>\text{diam}(X), a contradiction. Now let p0R[x0]p_{0}\in R[x_{0}] and let yR[p0]y\in R[-p_{0}]. Using that dis(R)π2\text{dis}(R)\leq\frac{\pi}{2}, we obtain

diam(X)2dX(x0,y)2(d𝕊1(p0,p0)π2)=π.\text{diam}(X)\geq 2d_{X}(x_{0},y)\geq 2\left(d_{\mathbb{S}^{1}}(p_{0},-p_{0})-\frac{\pi}{2}\right)=\pi.

We will now deduce the rest of the theorem. First note that for all xXx\in X we have dX(x,Π(x))π4d_{X}(x,\Pi(x))\leq\frac{\pi}{4}; indeed, as XX has length 5π4\leq\frac{5\pi}{4}, we have

dX(x,Π(x))=length([x,Π(x)])length(X)length([x,x+])5π4ππ4.d_{X}(x,\Pi(x))=\text{length}([x,\Pi(x)])\leq\text{length}(X)-\text{length}([x_{-},x_{+}])\leq\frac{5\pi}{4}-\pi\leq\frac{\pi}{4}.

And if dX(x,Π(x))=π4d_{X}(x,\Pi(x))=\frac{\pi}{4} for some xXx\in X, then all the inequalities above are equalities, so dX(x,x+)=πd_{X}(x_{-},x_{+})=\pi and length(X)=length([x,x+])+length([x,Π(x)])\text{length}(X)=\text{length}([x_{-},x_{+}])+\text{length}([x,\Pi(x)]), so the tree XX must be the union [x,x+][x,Π(x)][x_{-},x_{+}]\cup[x,\Pi(x)] and we would be done proving 4.10. We can therefore suppose dX(x,Π(x))<π4d_{X}(x,\Pi(x))<\frac{\pi}{4} for all xXx\in X, and we will obtain a contradiction.

For each p𝕊1p\in\mathbb{S}^{1} choose a point xpR[p]x_{p}\in R[p]. Note that we have dX(xp,xp)d𝕊1(p,p)π2=π2d_{X}(x_{p},x_{-p})\geq d_{\mathbb{S}^{1}}(p,-p)-\frac{\pi}{2}=\frac{\pi}{2} for all p𝕊1p\in\mathbb{S}^{1}. Moreover, Π(xp)Π(xp)\Pi(x_{p})\neq\Pi(x_{-p}) for all pp, because

dX(Π(xp),Π(xp))dX(xp,xp)dX(xp,Π(xp))dX(xp,Π(xp))>π2π4π4=0.d_{X}(\Pi(x_{-p}),\Pi(x_{-p}))\geq d_{X}(x_{p},x_{-p})-d_{X}(x_{p},\Pi(x_{p}))-d_{X}(x_{-p},\Pi(x_{-p}))>\frac{\pi}{2}-\frac{\pi}{4}-\frac{\pi}{4}=0. (5)

So, if we order the segment [x,x+][x_{-},x_{+}] by x>yx>y iff dX(x,x)>dX(x,y)d_{X}(x_{-},x)>d_{X}(x_{-},y), then for any p𝕊1p\in\mathbb{S}^{1} we have either Π(xp)>Π(xp)\Pi(x_{p})>\Pi(x_{-p}) or Π(xp)<Π(xp)\Pi(x_{p})<\Pi(x_{-p}).

Now, let (pt)t[0,π](p_{t})_{t\in[0,\pi]} be a geodesic segment of length π\pi in 𝕊1\mathbb{S}^{1}, so that pπ=p0p_{\pi}=-p_{0}. We can suppose w.l.o.g. that Π(xp0)<Π(xp0)\Pi(x_{p_{0}})<\Pi(x_{-p_{0}}). Thus, Π(xpπ)>Π(xpπ)\Pi(x_{p_{\pi}})>\Pi(x_{-p_{\pi}}).

Let s:=inf{t[0,π];Π(xp0)>Π(xp0)}[0,π]s:=\inf\{t\in[0,\pi];\Pi(x_{p_{0}})>\Pi(x_{-p_{0}})\}\in[0,\pi]. Suppose that Π(xps)>Π(xps)\Pi(x_{p_{s}})>\Pi(x_{-p_{s}}) (the case Π(xps)<Π(xps)\Pi(x_{p_{s}})<\Pi(x_{-p_{s}}) is similar). Then by definition of ss, there has to be some sequence (qn)n(q_{n})_{n} in 𝕊1\mathbb{S}^{1} convergent to psp_{s} and such that Π(xqn)<Π(xqn)\Pi(x_{q_{n}})<\Pi(x_{-q_{n}}) for all nn. By taking a subsequence, we can assume that (xqn)n,(xqn)n(x_{q_{n}})_{n},(x_{-q_{n}})_{n} converge when nn\to\infty to some points x,xXx_{\infty},x_{-\infty}\in X respectively, so that Π(x)Π(x)\Pi(x_{\infty})\leq\Pi(x_{-\infty}) and, as RX×𝕊1R\subseteq X\times\mathbb{S}^{1} is closed, we have xR[ps]x_{\infty}\in R[p_{s}] and xR[ps]x_{-\infty}\in R[-p_{s}].

So we have points xps,xR[ps]x_{p_{s}},x_{\infty}\in R[p_{s}] and xps,xR[ps]x_{-p_{s}},x_{-\infty}\in R[-p_{s}] with Π(xps)Π(xps)\Pi(x_{p_{s}})\geq\Pi(x_{-p_{s}}) and Π(x)Π(x)\Pi(x_{\infty})\leq\Pi(x_{-\infty}). Thus, we cannot have that both Π(xps)\Pi(x_{p_{s}}) and Π(x)\Pi(x_{\infty}) are strictly smaller than both Π(xps)\Pi(x_{-p_{s}}) and Π(x)\Pi(x_{-\infty}) or viceversa. Equivalently, there are points x,y,x{xps,xps,x,x}x,y,x^{\prime}\in\{x_{p_{s}},x_{-p_{s}},x_{\infty},x_{-\infty}\} such that Π(x)Π(y)Π(x)\Pi(x)\leq\Pi(y)\leq\Pi(x^{\prime}) and for some p{ps,ps}p\in\{p_{s},-p_{s}\} we have x,xR[p]x,x^{\prime}\in R[p] and yR[p]y\in R[-p] (this in turn implies that Π(x)<Π(y)<Π(x)\Pi(x)<\Pi(y)<\Pi(x^{\prime}): we cannot have equality by Equation 5). But then, we have a contradiction: as dis(R)=π2\text{dis}(R)=\frac{\pi}{2},

π2dX(x,x)=dX(x,Π(x))+dX(Π(x),Π(y))+dX(Π(y),Π(x))+dX(Π(x),x)=dX(x,y)+dX(y,x)2d(y,Π(y))>π2+π22π4=π2.\frac{\pi}{2}\geq d_{X}(x,x^{\prime})=d_{X}(x,\Pi(x))+d_{X}(\Pi(x),\Pi(y))+d_{X}(\Pi(y),\Pi(x^{\prime}))+d_{X}(\Pi(x^{\prime}),x^{\prime})\\ =d_{X}(x,y)+d_{X}(y,x^{\prime})-2d(y,\Pi(y))>\frac{\pi}{2}+\frac{\pi}{2}-2\frac{\pi}{4}=\frac{\pi}{2}.\qed

Now, let [x1,x2][x_{1},x_{2}] be the segment of length π4\frac{\pi}{4} we append to [x,x+][x_{-},x_{+}] to obtain XX, with x2[x,x+]x_{2}\in[x_{-},x_{+}]. We necessarily have dX(x0,x2)π4d_{X}(x_{0},x_{2})\leq\frac{\pi}{4} (as diam(X)=π\text{diam}(X)=\pi), and we may assume without loss of generality that dX(x,x2)dX(x+,x2)d_{X}(x_{-},x_{2})\leq d_{X}(x_{+},x_{2}). Thus we already have a very concrete description of XX, illustrated in Figure 4. Note that XX is determined up to isometry by dX(x0,x2)d_{X}(x_{0},x_{2}), so we will be done if we prove that x2=x0x_{2}=x_{0}, in which case the space XX is isometric to E.

Refer to caption
Fig. 4: Notation for the tree XX.

For each p𝕊1p\in\mathbb{S}^{1}, we will denote by Ip𝕊1I_{p}\subseteq\mathbb{S}^{1} the open interval centered at pp with radius π4\frac{\pi}{4}. Also, we divide the tree XX into two parts: X+:=[x0,x+]X_{+}:=[x_{0},x_{+}] and X=[x0,x][x0,x1]X_{-}=[x_{0},x_{-}]\cup[x_{0},x_{1}].

Claim 4.11.

Let pR[x0]p\in R[x_{0}]. Then R[p]{x,x1,x+}R[-p]\subseteq\{x_{-},x_{1},x_{+}\}, and exactly one of the following cases happens:

  1. Case 1.

    R[p]={x+}R[-p]=\{x_{+}\}. Then no point of IpI_{-p} can be related to any point of XX_{-}.

  2. Case 2.

    R[p]{x,x1}R[-p]\subseteq\{x_{-},x_{1}\}. Then no point of IpI_{-p} can be related to a point of X+X_{+}.

Proof.

There has to be some point yy in R[p]R[-p], because RR is a correspondence. As RR has distortion π2\frac{\pi}{2} and d𝕊1(p,p)=πd_{\mathbb{S}^{1}}(p,-p)=\pi, we have dX(x0,y)π2d_{X}(x_{0},y)\geq\frac{\pi}{2}, which implies y{x,x1,x+}y\in\{x_{-},x_{1},x_{+}\} (and y=x1y=x_{1} may only happen if dX(x0,x2)=π4d_{X}(x_{0},x_{2})=\frac{\pi}{4}).

If y=x+y=x_{+}, then for any p1Ipp_{1}\in I_{-p} and xR[p1]x\in R[p_{1}] we have:

dX(x0,x)\displaystyle d_{X}(x_{0},x) d𝕊1(p,p1)π2>π4.\displaystyle\geq d_{\mathbb{S}^{1}}(p,p_{1})-\frac{\pi}{2}>\frac{\pi}{4}.
dX(x+,x)\displaystyle d_{X}(x_{+},x) d𝕊1(p,p1)+π2<3π4.\displaystyle\leq d_{\mathbb{S}^{1}}(-p,p_{1})+\frac{\pi}{2}<\frac{3\pi}{4}.

But no point xXx\in X_{-} can satisfy both of these inequalities at the same time, thus, as we wanted, no point of IpI_{-p} can be related to any point of XX_{-}. The cases y=x1y=x_{1} and y=xy=x_{-} are similar. Also note that if x+R[p]x_{+}\in R[p], x1x_{1} or xx_{-} cannot be in R[p]R[p], because they are at distance >π2>\frac{\pi}{2} from x+x_{+} and dis(R)=π2\textup{dis}(R)=\frac{\pi}{2}. ∎

Claim 4.12.

We have R[x0]={p0,p0}R[x_{0}]=\{p_{0},p_{0}^{\prime}\} for some p0,p0𝕊1p_{0},p_{0}^{\prime}\in\mathbb{S}^{1} with d𝕊1(p0,p0)=π2d_{\mathbb{S}^{1}}(p_{0},p_{0}^{\prime})=\frac{\pi}{2} and such that R[p0]={x+}R[-p_{0}]=\{x_{+}\}, R[p0]{x,x1}R[-p_{0}^{\prime}]\subseteq\{x_{-},x_{1}\}.

Proof.

We first prove that there are points p0,p0R[x0]p_{0},p_{0}^{\prime}\in R[x_{0}] in Case 1. and Case 2. of 4.11 respectively. Suppose for contradiction that all the points of R[x0]R[x_{0}] are in Case 2. (the same argument works for Case 1.), so that all points in pR[x0]Ip\bigcup_{p\in R[x_{0}]}I_{-p} are not related to points of X+X_{+}. Note that any point y[x,x0]y\in[x_{-},x_{0}] with 0<dX(y,x0)<0.10<d_{X}(y,x_{0})<0.1 satisfies XBX(y,π2)X_{-}\subseteq B_{X}\left(y,\frac{\pi}{2}\right). So as dis(R)π2\text{dis}(R)\leq\frac{\pi}{2}, for any qR[y]q\in R[y] its antipodal q-q must be related to some point of X+X^{+}. Thus qpR[x0]Ipq\not\in\bigcup_{p\in R[x_{0}]}I_{p}. Letting yy approach x0x_{0}, and using that RX×𝕊1R\subseteq X\times\mathbb{S}^{1} is closed, we obtain that there is some point of R[x0]R[x_{0}] outside pR[x0]Ip\bigcup_{p\in R[x_{0}]}I_{p}, a contradiction.

So let p0,p0R[x0]p_{0},p_{0}^{\prime}\in R[x_{0}] satisfy Case 1. and Case 2. of 4.11 respectively. Then we necessarily have d𝕊1(p0,p0)π2d_{\mathbb{S}^{1}}(p_{0},p_{0}^{\prime})\geq\frac{\pi}{2}; if not, Ip0Ip0I_{-p_{0}}\cap I_{-p_{0}^{\prime}} would be nonempty, and its points could not be related to either XX_{-} or X+X_{+}. Now, as diam(R[x0])π2\textup{diam}(R[x_{0}])\leq\frac{\pi}{2}, R[x0]R[x_{0}] has to be contained in the geodesic segment [p0,p0][p_{0},p_{0}^{\prime}], and all points in R[x0]R[x_{0}] have to be at distance π2\geq\frac{\pi}{2} from either p0p_{0} or p0p_{0}^{\prime} for the same reasoning we just stated. So R[x0]={p0,p0}R[x_{0}]=\{p_{0},p_{0}^{\prime}\}, as we wanted. ∎

We now conclude the proof of 4.5. Let p0,p0p_{0},p_{0}^{\prime} be as in 4.12 and let p3,p4𝕊1p_{3},p_{4}\in\mathbb{S}^{1} be at the same distance from p0,p0p_{0},p_{0}^{\prime}, as in the figure below.

[Uncaptioned image]

Note that, as RR is closed in X×𝕊1X\times\mathbb{S}^{1} and all points of Ip0I_{-p_{0}} are related to some point of X+X_{+}, p4p_{4} has to be related to some point x4+X+x_{4}^{+}\in X_{+}. Similarly, p4p_{4} is related to some point x4Xx_{4}^{-}\in X_{-}. Note that dX(x0,x4+)+dX(x0,x4)=dX(x4+,x4)d_{X}(x_{0},x_{4}^{+})+d_{X}(x_{0},x_{4}^{-})=d_{X}(x_{4}^{+},x_{4}^{-}), so using that dis(R)=π2\textup{dis}(R)=\frac{\pi}{2}, we have

π2=d𝕊1(p4,p4)+π2dX(x4+,x4)=dX(x0,x4+)+dX(x0,x4)(d𝕊1(p0,p4)π2)+(d𝕊1(p0,p4)π2)=π4+π4=π2.\frac{\pi}{2}=d_{\mathbb{S}^{1}}(p_{4},p_{4})+\frac{\pi}{2}\geq d_{X}(x_{4}^{+},x_{4}^{-})=d_{X}(x_{0},x_{4}^{+})+d_{X}(x_{0},x_{4}^{-})\\ \geq\left(d_{\mathbb{S}^{1}}(p_{0},p_{4})-\frac{\pi}{2}\right)+\left(d_{\mathbb{S}^{1}}(p_{0},p_{4})-\frac{\pi}{2}\right)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}.

Thus the inequalities above are equalities, and dX(x0,x4+)=dX(x0,x4)=π4d_{X}(x_{0},x_{4}^{+})=d_{X}(x_{0},x_{4}^{-})=\frac{\pi}{4}.

Meaning that x4+x_{4}^{+} is the midpoint of [x0,x+][x_{0},x_{+}] and x4x_{4}^{-} is either the midpoint of [x0,x][x_{0},x_{-}] or some point in [x0,x1][x_{0},x_{1}]. Now, let x3R[p3]x_{3}\in R[p_{3}]. As d𝕊1(p3,p4)=πd_{\mathbb{S}^{1}}(p_{3},p_{4})=\pi and dis(R)=π2\textup{dis}(R)=\frac{\pi}{2}, we have dX(x3,x4+),dX(x3,x4)π2d_{X}(x_{3},x_{4}^{+}),d_{X}(x_{3},x_{4}^{-})\geq\frac{\pi}{2}. Now we divide in cases:

  • x4x_{4}^{-} is the midpoint of [x0,x][x_{0},x_{-}]. Then there are no points in XX at distance π2\geq\frac{\pi}{2} from both x4x_{4}^{-} and x4+x_{4}^{+}, unless x2=x0x_{2}=x_{0} and x3=x1x_{3}=x_{1}, proving 4.5.

  • x4x_{4}^{-} is in [x0,x1][x_{0},x_{1}]. If dX(x0,x2)=π4d_{X}(x_{0},x_{2})=\frac{\pi}{4}, then again x4=x2x_{4}^{-}=x_{2} is the midpoint of the segment [x0,x][x_{0},x_{-}], so we are back to the previous case. If dX(x0,x2)<π4d_{X}(x_{0},x_{2})<\frac{\pi}{4}, then R[p0]={x}R[-p_{0}^{\prime}]=\{x_{-}\} by the proof of 4.11. In this case we also have x2=x0x_{2}=x_{0}: if not, the inequalities dX(x3,x4+),dX(x3,x4)π2d_{X}(x_{3},x_{4}^{+}),d_{X}(x_{3},x_{4}^{-})\geq\frac{\pi}{2} would imply that x3x_{3} is at distance <π4<\frac{\pi}{4} from xx_{-}, which contradicts the facts that dis(R)=π2\textup{dis}(R)=\frac{\pi}{2} and d𝕊1(p3,p0)=3π4d_{\mathbb{S}^{1}}(p_{3},-p_{0})=\frac{3\pi}{4}.

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